Problem: Simplify and expand the following expression: $ \dfrac{3}{5p + 15}+ \dfrac{1}{3p - 21}- \dfrac{2}{p^2 - 4p - 21} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{3}{5p + 15} = \dfrac{3}{5(p + 3)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{1}{3p - 21} = \dfrac{1}{3(p - 7)}$ We can factor the quadratic in the third term: $ \dfrac{2}{p^2 - 4p - 21} = \dfrac{2}{(p + 3)(p - 7)}$ Now we have: $ \dfrac{3}{5(p + 3)}+ \dfrac{1}{3(p - 7)}- \dfrac{2}{(p + 3)(p - 7)} $ The least common multiple of the denominators is: $ 15(p + 3)(p - 7)$ In order to get the first term over $15(p + 3)(p - 7)$ , multiply by $\dfrac{3(p - 7)}{3(p - 7)}$ $ \dfrac{3}{5(p + 3)} \times \dfrac{3(p - 7)}{3(p - 7)} = \dfrac{9(p - 7)}{15(p + 3)(p - 7)} $ In order to get the second term over $15(p + 3)(p - 7)$ , multiply by $\dfrac{5(p + 3)}{5(p + 3)}$ $ \dfrac{1}{3(p - 7)} \times \dfrac{5(p + 3)}{5(p + 3)} = \dfrac{5(p + 3)}{15(p + 3)(p - 7)} $ In order to get the third term over $15(p + 3)(p - 7)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{2}{(p + 3)(p - 7)} \times \dfrac{15}{15} = \dfrac{30}{15(p + 3)(p - 7)} $ Now we have: $ \dfrac{9(p - 7)}{15(p + 3)(p - 7)} + \dfrac{5(p + 3)}{15(p + 3)(p - 7)} - \dfrac{30}{15(p + 3)(p - 7)} $ $ = \dfrac{ 9(p - 7) + 5(p + 3) - 30} {15(p + 3)(p - 7)} $ Expand: $ = \dfrac{9p - 63 + 5p + 15 - 30}{15p^2 - 60p - 315} $ $ = \dfrac{14p - 78}{15p^2 - 60p - 315}$